Counting statistics of random events: A
tutorial
This tutorial presents derivations of some results from the theory of the statistics of random events
that are studied in counting experiments.
The statistics of counting random events falls under the rubric of what statisticians call “point
processes”. A point process in time is one where a particular type of event is classified only
according to a single point on a time line. Some examples of phenomena that could generate time
points are the births of babies in a particular hospital, the passage of cars past a counting gate on
a toll bridge, or the arrival of particles from outer space at a detector. Such point processes may be
further categorized as “Poisson processes” if the occurrence of any particular point is completely
independent of the occurrence of any other point, as is the case in our cosmic ray experiment. If
the criterion of independence is not satisfied, the point process may be more generally known as
a “renewal process”, a name which is derived from the study of part failures and replacements in
industrial contexts (e.g., a new light bulb is not installed until the old one burns out) [4].
This section presents in a tutorial fashion the derivation of some important formulas in the theory
of Poisson processes. These results are neither new nor little known, especially among statisticians.
The goal is to show derivations that a student with the mathematical background of first-year
calculus can follow and believe. These are “physicist’s proofs” where a plausible argument is
chosen over rigorous demonstration. Suggested exercises are given to help the student cement his
or her understanding.
The raw data for our statistical investigations comes in the form of a random pulse train, as
illustrated in Fig. 1. The pulse width T w is important only insofar as it determines the maximum
rate of pulses that may be represented by the pulse train, since pulses which occur more frequently
than 1/Tw cannot be resolved. The occurrence time of each pulse, denoted as t 1 , t2 , etc., as
measured from some (arbitrary) starting time, is marked at the leading edge of each pulse.
t1
t2
t3
t4
t5
t6
t7
T45
Tw
T46
Figure 1: A random pulse train, as might be seen on an oscilloscope at a particular instant.
This example shows negative-going (“fast NIM”) pulses typical of those used in nuclear counting
experiments.
If the pulse train is being created by the detection of decays from a long-lived radioactive nuclide or
by cosmic rays, experiments show that although the time between specific pulses T ij is completely
unpredictable, the time intervals do converge to a well-defined average τ in the following sense:
If one counts very many pulses (i.e., thousands) N over a long time interval T , then τ ≈ T /N .
More specifically, if one estimates τ by counting N pulses over number of trials, the variation of the
1
calculated
τ ’s from these trials will cluster about each other with a fractional standard deviation
√
of 1/ N . (Later we will strengthen this assertion.) We can define an average rate of pulses r by
the relation r = 1/τ .
We are interested in two problems:
1. What is the distribution of intervals between successive pulses in a given pulse train? More
generally, what is the distribution of scaled intervals: intervals between two pulses separated
by an arbitrary number of pulses in between?
2. What is the distribution of counts within a succession of fixed-length periods?
To make these problems clear, we remind the reader of a few basic definitions. A distribution is the
collected result of many trials of a particular experiment. Each trial produces a value of a random
variable. For example, in the distribution of counts, one trial consists of adding up the number of
counts that happen in a particular time period. This number is the value of the random variable
N (t), where t denotes the time length of the period. The collection of numbers N (t i ), where i is
an index denoting each trial period, is the distribution. Since N (t) can only take on non-negative
integer values, the count distribution is discrete. The random variable in the interval distribution
is the time between pulses Tij . Since it can take on any nonzero value, the interval distribution
is continuous. Distributions are often presented as histograms (vertical bar graphs), in which the
horizontal axis represents the value (or range of values, in the case of a continuous distribution) of
the random variable and the height of the bars represents how often that value occurs.
The theoretical problem is the derivation of probability distribution functions: functions that give
the likelihood of finding a particular value or range of values of the random variable. We will
address the theoretical problem by making a basic assumption about our pulse train and invoking
two rules. The rules are from the theory of probabilities [1]. Assume that two events of a similar
type A and B follow from the same proximate cause. The probabilities of these events P (A) and
P (B) are subject to these two rules:
I. The rule of compound probabilities. If the two events A and B are mutually exclusive,
then the probability of having either event P (A or B) is given by the sum P (A) + P (B).
II. The rule of independent probabilities. If the two events A and B are independent
of each other, then the probability of having both events occur P (A and B) is given by
the product P (A) × P (B).
Simple examples of these two rules are illustrated by considering rolls of dice. A die has six
numbered sides, and when it is thrown any one number will be equally likely to land face up: the
probability of any number is 1/6. If we ask, “What is the probability to roll an even number?”, we
are invoking rule I, since the roll of any number excludes the roll of any other number. Thus,
P (even) = P (2) + P (4) + P (6) =
1
1 1 1
+ + = .
6 6 6
2
If we ask, “What is the probability to roll two 1s in a row?”, we are invoking rule II, since each
successive roll of the die is independent of any previous roll. Thus
P (1 & 1) = P (1) × P (1) =
2
1
1 1
× =
.
6 6
36
If we ask, “What is the probability of rolling only one 1 in two rolls of the die?”, we want to know
the probability of a 1 on the first roll and not 1 on the second or a 1 on the second roll and not
on the first. Thus, since successive rolls of the dice are independent events and the two possible
desired outcomes are mutually exclusive,
P (only one 1 in 2 rolls) = P (1; 1st) × P ∗ (1; 2nd) + P ∗ (1; 1st) × P (1; 2nd)
1 5
5 1
=
× + ×
6 6
6 6
10
,
=
36
where P ∗ is the compliment of P , that is, it is the probability of not obtaining the specified outcome.
By convention, probability distributions are normalized: the random variable must take on one of
its possible values, and since any value X either does or does not occur, P (X) + P ∗ (X) = 1, by
rule I.
The basic assumption we make about our data is
Any infinitesimally small interval of time dt is equally likely to contain a pulse. The
probability of finding a pulse in dt is given by r dt.
For this assumption to make sense, r dt must be much smaller than one, and the probability of
finding two pulses in dt must be negligible. As long as r is a constant, these restrictions can be
assured by making dt small enough.
We will first apply the rules we have stated to a real problem in particle physics: random coincidences.
Rate of random coincidences
The use of coincidence counting is pervasive in elementary particle experiments for one big reason:
noise. For example, if you set up a scintillator paddle to count the passage of cosmic-ray muons,
most of the counts you record will be due to non-muon events such as electrical noise, low-level
background radiation, etc. Fortunately, muons have fairly high energy, so that if they pass through
a thin scintillator paddle they will continue out the other side and can be detected by a second
paddle nearby. Such an event produces two pulses, one in each detector, at (nearly) the same time.
By using a logic gate, one can force the counting electronics to record only those pulses which occur
in coincidence. A schematic of this setup is shown in Fig. 2. In order for the coincidence technique
to work profitably, the rate of random coincidences must be appreciably lower than the rate of
“true” coincidences. We can use our results to predict this rate.
Let the rate of random pulses recorded by detector A be be rA and the rate of random pulses
recorded by detector B be rB . Each time a pulse from either A or B enters the gate, the gate is
triggered: it “opens” for a short time interval T called the resolving time, meaning that if a pulse
from the other detector occurs during T , then a coincidence pulse is produced at the output. Since
rA and rB are the average rates of random pulses, the rate of coincidences will also be random (in the
absence of any true coincidence events). Let R 2 be this rate; more specifically, we let R 2 dt be the
probability that a random coincidence event between 2 detectors will occur within the infinitesimal
time interval dt.
3
(a)
(b)
Muon path
A
A pulses
B
B pulses
DISC
COINCIDENCE GATE
B
DISC
A&B coincidence
A&B
T
A
Figure 2: (a) A two-paddle coincidence setup. Variable height detector pulses are discriminated
and turned into digital pulses. The coincidence gate produces a pulse when two input pulses overlap
within a gate time T . (b) Example of two pulse trains showing one coincidence.
By rules I and II R2 dt is given by the following construction:



Probability of the gate


 
R2 dt =  being triggered by
×

detector A in dt.


Probability of detector
B delivering a pulse
within resolving time
T of the trigger time.

Probability of the gate

 

+  being triggered by
×

detector B in dt.





Probability of detector
A delivering a pulse
within resolving time
T of the trigger time.



 .

In symbolic form, this equation reads
R2 dt = rA dt × rB T + rB dt × rA T .
Typically the resolving time T is much shorter than the average rates rA and rB so that rT 1.
In this case, we may approximate the probability of finding a pulse in the finite time T by using
the infinitesimal probability: r dt ≈ rT . So the rate of 2-fold random coincidences R 2 is given by
R2 = 2rA rB T .
(1)
Exercise 1 Show that if you added a third paddle C to the setup, the rate of random 3-fold coincidences R3 would be equal to 3rA rB rC T 2 .
Distribution of intervals
Now, let’s return to our main questions. We first calculate the probability that there will be no
pulse in a finite time interval T . Since we assume that a pulse may occur at any instant with equal
4
T
Possible pulses
∆Τ=Τ/Ν
N small intervals
Figure 3: Construction used in calculating the probability of zero pulses in a finite time T .
probability, it does not matter where along the time line we start our clock. Since our assumption
involves the infinitesimal interval dt, we divide T into small N small intervals ∆T , with an eye
toward taking the limit of small ∆T . This construction is shown in Fig. 3. In any interval of length
∆T , when this interval becomes small, the probability of finding a pulse is r∆T , so by Rule 1, the
probability of not finding a pulse is 1 − r∆T . Since the probability of finding a pulse in any small
interval is independent of the probability of finding a pulse in any other interval, the probability of
not finding a pulse in T , P0 (T ) is, by rule II, the product of the probabilities for all of the N small
intervals:
P0 (T ) ≈ (1 − r∆T )N .
(2)
If we write ∆T = T /N , the limit of infinitesimal ∆T is found by letting N get large, and in this
limit, the approximation tends to equality. Thus,
P0 (T ) = lim
N →∞
rT
1−
N
N
= e−rT .
(3)
The final step in this unsurprising result can be proved by taking the logarithm of both sides of
the equation, since ln(1 + x) ≈ x for small x.
We now use the result of Eq. (3) to derive the probability density function of 1-pulse intervals,
I1 (t). We define this continuous probability density function as follows: I 1 (t) dt is the probability
of finding a 1-pulse interval of length between t and t + dt. This means the probability of finding a
pulse during the infinitesimal time interval dt after a time interval of length t following a previous
pulse during which there have been no pulses. This situation is illustrated by Fig. 4.
If we pick a pulse in the train, and assign its time as t = 0, then we obtain I 1 (t) by the following
construction from rule II:




Probability of finding
Probability of finding

 

I1 (t) dt =  zero pulses between 0  ×  a pulse between t and 
and t
t + dt
= P0 (t) × r dt
= re−rt dt .
Thus
I1 (t) = re−rt .
(4)
This result is known as the exponential distribution and has a couple of interesting features. First,
the condition that each infinitesimal interval dt is equally likely to contain a pulse leads to the
5
t
dt
NO PULSE
PULSE?
Figure 4: The construction leading to I 1 (t) dt asks how likely is a pulse to occur in dt after a
previous pulse.
result that shorter intervals are much more common than longer ones. This is a vindication of the
colloquial rule that “disasters come in threes”: completely random events tend to cluster together
in time. Perhaps more practical is the prediction that one may find the average rate r in a counting
experiment by plotting the experimental interval distribution on a semi-log scale: both the slope
and intercept will give r (after a some arithmetic).
Exercise 2 (a) Show that I1 (t) is normalized, that is, prove 0∞ I1 (t) dt =R1. (b) Show that the
average 1-pulse interval τ1 is equal to 1/r. The average interval is given by 0∞ tI1 (t) dt.
R
The situation for the distribution of 2-pulse intervals is a bit more complicated. We want the
likelihood of finding an interval of a particular length t between two pulses that contains a single
pulse anywhere inside at t0 < t. One such possibility is shown in Fig. 5. The (differential) probability
of finding an interval like this is given by rule II:
dP (t, t0 ) =

Probability of
 finding 0


 pulses between
0 and t0


 
 
×
 
Probability of
finding a pulse
between t0 and
t0 + dt0


 
 
×
 
Probability of
finding 0
pulses between
t0 and t


 
 
×
 
Probability of
finding a pulse
between t and
t + dt





= P0 (t0 ) × r dt0 × P0 (t − t0 ) × r dt .
This construction is nothing more than the probability of finding two 1-pulse intervals of particular
lengths. But since the intermediate pulse can happen anywhere in the interval, by rule I we must
sum the individual probabilities over all values of t 0 between 0 and t. Hence, we see that
I2 (t) dt =
Z
=
Z
=
t
I1 (t0 )I1 (t − t0 ) dt0
0
t
re
−rt0
re
−r(t−t0 )
0
2 rt
r e
Z
t
dt
0
0
dt
0
dt
dt
dt
= r(rt)e−rt dt .
So the distribution function for 2-pulse intervals is
I2 (t) = r(rt)e−rt .
6
(5)
0
NO PULSES
t’
t dt
PULSE?
PULSE!
Figure 5: Construction used in calculating the probability a 2-pulse interval of length t.
Likewise, we may find the distribution for 3-pulse intervals I 3 (t) by multiplying I2 (t0 ) by I1 (t − t0 )
an integrating over all t0 < t. This process leads to a general recursion relation
In (t) =
Z
t
0
In−1 (t0 )I1 (t − t0 ) dt0 .
(6)
When applied to the exponential distribution the formula gives
In (t) = r
(rt)(n−1) e−rt
.
(n − 1)!
(7)
Equation 7 is known as the (integer) gamma distribution or Erlang(ian) distribution (after the
Danish engineer A. K. Erlang who used it to characterize telephone networks).
Exercise 3 Convince yourself of the (n − 1)! factor in Eq. (7) by using I 3 (t) and I1 (t) to show that
I4 (t) = r
(rt)3 e−rt
1·2·3
Exercise 4 Show that the average n-pulse interval τ n is equal to n/r, but that the most probable
interval Tn,peak is given by (n − 1)/r. The most probable interval
is the interval at the peak of the
R
distribution, and is found by evaluating dI n (t)/dt = 0. Hint: 0∞ xn e−x dx = n!.
Graphs of In (t) are shown in Fig. 6(a) for a few values of n. A couple of things are notable. First,
the probability of finding zero length intervals for n > 1 is zero. This makes sense in that there
must be at least one pulse within each such n-pulse interval, and even a single pulse takes time.
More interestingly, the relative width of the peak of I n (t) shrinks as n increases. This is shown most
clearly by scaling In (t) so that the functions all have the same mean value (1/r) while maintaining
the normalized (unit) area, as is shown in Fig. 6(b). What this means is that for larger n, the
collection of intervals becomes more uniform.
7
1
1.5
(b)
(a)
0.8
n =10
n=1
1
n =5
(1/r)In(t)
(n/r)In(t)
0.6
0.4
n=2
n=3
0.2
0
0
1
2
3
n =2
0.5
n =1
n=4
4
rt
n=5
5
7
6
0
8
0
1
2
rt/n
3
4
Figure 6: The n-pulse interval density functions, after Knoll [2, p. 98]. (a) (1/r)I n (t) for n from 1
to 5. Notice that the functions peak at rt = n − 1. (b) (1/r)I n (t) scaled by the transformations
t → t/n and (1/r)In (t) → (n/r)In (t) to show that as n increases, the function becomes more
sharply peaked around the mean value τ n = n/r.
Exercise 5 Evaluate the variance σn2 of In (t). The variance is given by the formula
σn2 =
Z
∞
0
(t2 − τn2 )In (t) dt .
√
Then show that the fractional standard deviation in τ n , σn /τn is equal to 1/ n.
Distribution of counts
Up to now, we have been framing our investigation thusly, “Given a fixed number of counts, how
are the intervals that contain exactly this number distributed?” In other words, we are using the
count number as a parameter in our problem, and letting the time be the variable. But frequently,
our experimental situation is different: the independent parameter is the time period, and the
dependent variable is the number of counts that occur within that time period. This is a related,
but not identical question. Most obviously, as stated in the introduction, a time interval can take
on any value, but a count number can only take integer values; the distribution of interval lengths
is a continuous distribution, whereas the distribution of counts is a discrete distribution.
The relationship between the two distributions can be stated plainly: A period of a given length
containing q counts, Tq , has fewer than n counts if and only if the interval containing n counts, T n ,
is longer than Tq . That is,
q < n if and only if Tn > Tq .
This suggests the following interpretation of what has been derived so far. The probability of
finding an n-count interval of between length T and T + dt is the same as the probability of finding
8
Table 1: A comparison of the two types of distributions that are contained in the “Poisson” formula.
Note σE2 = τn2 /n and σP2 = n.
Type:
Variable:
Mean:
Variance:
Erlang
Continuous
t
τn = n/r
σE2 = n/r 2
Poisson
Discrete
n
n = rt
σP2 = rt
exactly n − 1 counts in an interval of length T times the probability of finding one more count in
T to T + dt. Stated in terms of our formulas:
In (T ) dt =
(rT )(n−1) e−rT
× r dt = P (n − 1; rT ) × r dt .
(n − 1)!
(8)
Thus, we identify P (n − 1; rT ) as none other than the Poisson distribution giving the probability of
finding n−1 events given that the average number of events is rT . This fact is used in the derivations
of the interval distribution function given by Knoll [2] and Melissinos [3]. Most derivations of the
Poisson distribution are based on taking a particular limit of the discrete binomial distribution.
The derivation given here follows a more direct route from the basic rules of probability as applied
to random pulse trains.
As is well known, the Poisson distribution is a distribution where the mean is equal to the variance;
only one parameter is needed to define the whole distribution. This is not true of the Erlang
distribution; for a given rate constant r and count number n, the mean interval time is n/r but the
variance is n/r 2 . A comparison of the two distributions given by our formula is given in Table 1.
References
[1] Wannier, Gregory H., Statistical Physics, (Dover, Mineola NY, 1987), p. 29.
[2] Knoll, Glenn F., Radiation Detection and Measurement, 2nd edition, (John Wiley, New York,
1989), pp. 96–99.
[3] Melissinos, Adrian C., and Jim Napolitano, Experiments in Modern Physics, 2nd edition (Academic Press, New York, 2003), pp. 401–404, 470–473.
[4] Cox, D. R., Renewal Theory, (Meuthen and Co. Science Paperbacks, Butler & Tanner Ltd.,
1967).
Prepared D. Pengra
counting_stats_tutorial_b.tex -- Updated 29 April 2008
9